What is the solution for (p*m-p^2)^{-1}*dp=k*dt ?

The solution for (p*m-p^2)^{-1}*dp=k*dt is t=-(-ln(p/m)+ln(p/m-1))/(mk)+c_{1}

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(pm-p^2)^{-1}dp=k*dt

(p \cdot m - p^{2})^{- 1} \cdot d p = k \cdot d t

t = - \frac{- ln ( \frac{p}{m} ) + ln ( \frac{p}{m} - 1 )}{mk} + c_{1}

Solution steps

(p \cdot m - p^{2})^{- 1} \cdot d p = k \cdot d t

Let

t

be the dependent variable. Divide by

d p

(p m - p^{2})^{- 1} = k \frac{d t}{d p}

Substitute

\frac{d t}{d p}

with

t^{'} (p)

(p m - p^{2})^{- 1} = k t^{^{'}} (p)

Isolate

t^{'} (p) : t^{'} (p) = \frac{1}{k ( m p - p ^{2} )}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

t = \int \frac{1}{k ( m p - p ^{2} )} d p

\int \frac{1}{k ( m p - p ^{2} )} d p = - \frac{1}{km} (- ln (\frac{p}{m}) + ln (\frac{p}{m} - 1)) + c_{1}

t = - \frac{1}{km} (- ln (\frac{p}{m}) + ln (\frac{p}{m} - 1)) + c_{1}

Simplify

t = - \frac{- ln ( \frac{p}{m} ) + ln ( \frac{p}{m} - 1 )}{mk} + c_{1}

\frac{d}{d x} (9 x^{5} cos (x))

\frac{d}{d x} (x + x^{5})

f (x) = 11 - x + 4 ln (2 x - 1), at x = 1

\frac{1}{1 + x ^{4}}

f (x) = \frac{x}{1 + x ^{2}}

What is the solution for (p*m-p^2)^{-1}*dp=k*dt ?
The solution for (p*m-p^2)^{-1}*dp=k*dt is t=-(-ln(p/m)+ln(p/m-1))/(mk)+c_{1}